NCERT Solutions for Class 7 Maths Chapter 3 - Data Handling

First, we have to arrange the given marks in ascending order.

= 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9

Now, we will draw the frequency table of the given data.

(i) By observing the table clearly, the highest number among the given data is 9.

(ii) By observing the table clearly, the lowest number among the given data is 1.

(iii) We know that Range = Highest value – Lowest value

= 9 – 1

= 8

(iv) Now, we have to calculate Arithmetic Mean,

Arithmetic mean = (Sum of all observations)/ (Total number of observations)

Then,

Sum of all observation = 1 + 2 + 2 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 7 + 7

+ 8 + 9

= 100

Total Number of Observations = 20

Arithmetic mean = (100/20)

= 5

 

The first five Whole numbers are 0, 1, 2, 3, and 4.

Mean = (Sum of first five whole numbers)/ (Total number of whole numbers)

Then,

Sum of five whole numbers = 0 + 1 + 2 + 3 +4

= 10

Total Number of whole numbers = 5

Mean = (10/5)

= 2

∴ The mean of the first five whole numbers is 2.

 

Mean score = (Total runs scored by the cricketer in all innings)/ (Total number of innings

played by the cricketer)

Total runs scored by the cricketer in all innings = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100

= 400

Total number of innings = 8

Then,

Mean = (400/8)

= 50

∴ The mean score of the cricketer is 50.

 

Let us assume the heights (in cm) of 10 students in our class be

= 130, 132, 135, 137, 139, 140, 142, 143, 145, 148

By observing the above-mentioned values, the highest value is = 148 cm

By observing the above-mentioned values, the lowest value is = 130 cm

Then,

Range of Heights = Highest value – Lowest value

= 148 – 130

= 18 cm

 

(i) A’s average number of points scored per game = Total points scored by A in 4 games/

Total number of games

= (14 + 16 + 10 + 10)/ 4

= 50/4

= 12.5 points

(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played only 3 games.

(iii) B played in all four games, so we will divide the total points by 4 to find out the mean.

Then,

Mean of B’s score = Total points scored by B in 4 games/ Total number of games

= (0 + 8 + 6 + 4)/ 4

= 18/4

= 4.5 points

(vi) Now, we have to find the best performer among the 3 players.

So, we have to find the average points of C = (8 + 11 + 13)/3

= 32/3

= 10.67 points

By observing, the average points scored A is 12.5, which is more than B and C.

Clearly, we can say that A is the best performer among the three.

 

First, we have to arrange the marks obtained by a group of students in a science test in ascending order.

= 39, 48, 56, 75, 76, 81, 85, 85, 90, 95

(i) The highest marks obtained by the student = 95

The lowest marks obtained by the student = 39

(ii) We know that Range = Highest marks – Lowest marks

= 95 – 39

= 56

(iii) Mean of Marks = (Sum of all marks obtained by the group of students)/

(Total number of marks)

= (39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95)/ 10

= 730/10

= 73

 

Mean enrolment = Sum of all observations / Number of observations

= (1555 + 1670 + 1750 + 2013 + 2540 + 2820)/ 6

= (12348/6)

= 2058

∴ The mean enrolment of the school for this given period is 2058.

 

(i) Range of rainfall = Highest rainfall – Lowest rainfall

= 20.5 – 0.0

= 20.5 mm

(ii) Mean of rainfall = Sum of all observations / Number of observations

= (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)/ 7

= 41.3/7

= 5.9 mm

(iii) We may observe that for 5 days, i.e. Monday, Wednesday, Thursday, Saturday and Sunday, the rainfall was less than the average rainfall.

 

First, we have to arrange the given data in ascending order.

= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151

(i) The height of the tallest girl is 151 cm.

(ii) The height of the shortest girl is 128 cm.

(iii) Range of given data = Tallest height – Shortest height

= 151 – 128

= 23 cm

(iv) Mean height of the girls = Sum of the height of all the girls / Number of girls

= (128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150

+ 151)/ 10

= 1414/10

= 141.4 cm

(v) 5 girls have heights more than the mean height (i.e. 141.4 cm).

 

Arranging the given scores in ascending order, we get

5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

Mode

Mode is the value of the variable which occurs most frequently.

Clearly, 20 occurs the maximum number of times.

Hence, the mode of the given sores is 20

Median,

The value of the middle-most observation is called the median of the data.

Here, n = 15, which is odd.

Where n is the number of students.

∴ median = value of ½ (n + 1)th observation

= ½ (15 + 1)

= ½ (16)

= 16/2

= 8

Then, the value of the 8th term = 20

Hence, the median is 20.

Yes, both values are the same.

 

Arranging the runs scored in a cricket match by 11 players in ascending order, we get

6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120

Mean

Mean of the given data = Sum of all observations/Total number of observations

= (6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120)/ 11

= 429/11

= 39

Mode

Mode is the value of the variable which occurs most frequently.

Clearly, 15 occurs the maximum number of times.

Hence, the mode of the given sores is 15.

Median,

The value of the middle-most observation is called the median of the data.

Here, n = 11, which is odd.

Where n is the number of players.

∴ median = value of ½ (n + 1)th observation

= ½ (11 + 1)

= ½ (12)

= 12/2

= 6

Then, the value of the 6th term = 15

Hence, the median is 15.

No, these three are not the same.

 

Arranging the given weights of 15 students of a class in ascending order, we get

32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50

(i) Mode and Median

Mode

Mode is the value of the variable which occurs most frequently.

Clearly, 38 and 43 both occur 3 times.

Hence, the mode of the given weights is 38 and 43.

Median

The value of the middle-most observation is called the median of the data.

Here, n = 15, which is odd.

Where n is the number of students.

∴ median = value of ½ (n + 1)th observation

= ½ (15 + 1)

= ½ (16)

= 16/2

= 8

Then, the value of the 8th term = 40

Hence, the median is 40.

(ii) Yes, there are 2 modes for the given weights of the students.

 

Arranging the given data in ascending order, we get

= 12, 12, 13, 13, 14, 14, 14, 16, 19

Mode

Mode is the value of the variable which occurs most frequently.

Clearly, 14 occurs the maximum number of times.

Hence, the mode of the given data is 14.

Median

The value of the middle-most observation is called the median of the data.

Here n = 9, which is odd.

Where n is the number of students.

∴ median = value of ½ (9 + 1)th observation

= ½ (9 + 1)

= ½ (10)

= 10/2

= 5

Then, the value of the 5th term = 14

Hence, the median is 14.

 

(i) The statement given above is true.

Because mode is the value of the variable which occurs most frequently in the given data.

Hence, the mode is always one of the numbers in the data.

(ii) The statement given above is false.

Because mean may or may not be one of the numbers in a data.

(iii) The statement given above is true.

Because the median is the value of the middle-most observation in the given data while arranged in ascending or descending order.

Hence, the median is always one of the numbers in a data

(iv) Mean = Sum of all given observations/ number of observations

= (6 + 4 + 3 + 8 + 9 + 12 + 13 + 9)/8

= (64/8)

= 8

Hence, the given statement is false.

 

The bar graph represents the pets owned by the students.

(a) From the bar graph, the most popular pet is a cat. It is owned by 10 students out of 12 students.

(b) From the bar graph, 8 students have a dog as a pet out of 12 students.

 

(i) By observing the bar graph,

175 books were sold in the year 1989.

475 books were sold in the year 1990.

225 books were sold in the year 1992.

(ii) By observing the bar graph,

475 books were sold in the year 1990.

225 books were sold in the year 1992.

(iii) By observing the bar graph,

In the years 1989 and 1992, the number of books sold was less than 250.

(iv) By observing the bar graph, we can conclude that,

The number of books sold in the year 1989 is about 1 and ¾th part of 1 cm.

WKT, Scale is taken as 1 cm = 100 books

= 100 + (¾ × 100)

= 100 + (3 × 25)

= 100 + 75

= 175

 

(a) We will take the scale as 1 unit = 10 children because we can represent a bigger and clear difference among the number of students in class 7th and class 9th.

(b) (i) Class 5th has the maximum number of children, i.e., 135, and class 7th has the minimum number of children, i.e., 95.

(ii) The total number of students in class 6th is 120, and the total number of students in class 8th is 100.

Then,

Ratio between the number of students in class 6th and 8th,

= (120/100)

= 6/5

= 6: 5

 

(i) By observing the double bar graph, there was a maximum mark increase in the Maths subject. So, the child has improved his performance in Maths.

(ii) By observing the double bar graph, the improvement was the least in S. Science.

(iii) By observing the double bar graph, the performance In Hindi has gone down.

 

(i) The fig above is the double bar graph, which represents the people who prefer either watching or participating in different sports. By observing the double bar graph, we came to conclude that most people like watching and participating in cricket, while the least number of people like watching and participating in athletics.

(ii) By observing the double bar graph, we came to conclude that the people who like watching and participating in cricket are the tallest among all the bars. So, cricket is the most popular sport.

(iii) By observing the double bar graph, we came to conclude that watching sports has more preference, rather than participating in sports.

 

(i) By observing the double bar graph, we came to conclude that Jammu has the largest difference in the minimum and maximum temperature on 20.6.2006.

(ii) By observing the double bar graph, we came to conclude that Jammu is the hottest city and Bangalore is the coldest city.

(iii) By observing the double bar graph, Bangalore and Jaipur, Bangalore and Ahmedabad, for Bangalore, the maximum temperature was 28oC, while the minimum temperature of both Ahmadabad and Jaipur was 29oC.

(iv) By observing the double bar graph, Mumbai has the least difference between its minimum and maximum temperature.

 

(i) It is certain to happen.

(ii) It can happen but not certain.

(iii) It is impossible because there are only six faces on a die, marked as 1, 2, 3, 4, 5, and 6 on it.

(iv) It can happen but not certain.

(v) It can happen but not certain.

A coin has two faces, one is the Head, and the other one is the Tail.

Now, one team can choose either Head or Tail.

The probability of our team starting the first= Number of favourable outcomes/

Number of possible outcomes

= ½

(i) From the question, it is given that

There are 6 marbles in the box with numbers from 1 to 6 marked.

The probability of drawing a marble with the number 2 = Number of favourable outcomes/

Number of possible outcomes

= (1/6)

(ii) From the question, it is given that

There are 6 marbles in the box with numbers from 1 to 6 marked.

The probability of drawing a marble with the number 5 = Number of favourable outcomes/

Number of possible outcomes

= (1/6)